y(x)=4/(1+tan^2x)-3

y(x)=4/(1+tan^2x)-3

Hint: replace tan^2x with sin^2x/cos^2x and see where it takes you.

Hint: replace tan^2x with sin^2x/cos^2x and see where it takes you.

Didn't help

didnt help

What did you get? After making this substitution try multiplying numerator and denominator by cos^2x.

I know to identities this has just thrown me to the deep end

I got it down to y(x)=4cos^2/(1+(1/cos^2x)-3

no one can help?. I put it in wolfram and it gives y(x)=1-4sin^2(x)

any help on how to get to there

i got it down to y(x)=4cos^2/(1+(1/cos^2x)-3

I think you made a mistake - what happened to the sin^2x term in the denominator?

Substituting \tan^2x = \sin^2x/ \cos^2x:


\frac {4} { 1 + \tan^2x} - 3 = \frac 4 {1 + \frac {\sin^2x}{\cos^2x}} - 3


Now multiply both numerator and denomninator by cos^2x:


\frac {4 \cos ^2x}{\cos^2x + \sin^2x} -3


You should recognize that \cos^2x + \sin^2x=1, so this becomes 4\cos^2x - 3. Now if you want you can subsitite \cos^2x = 1- \sin^2x to get the Wolfram solution.

yeah my mistake was when I multiply sin^2x/cos^2x by cos^2x would not that cancel out the cos on the denominator?

what program you use to write your equations?

yeah my mistake was when i multiply sin^2x/cos^2x by cos^2x would not that cancel out the cos on the denominator?

yes it does, leaving the sin^2x term..



what program you use to write your equations?

LaTeX - it allows for embedding math equations into questions and responses. You can learn about it here:
https://www.askmehelpdesk.com/math-sciences/how-technical-scientific-documentation-formulas-50415.html

yeah agreed so for my information in the denominator

the sin^2x/cos^2x was multiply by cos^2x/cos^2x to leave sin^2x which makes it

4/1+sin^2x

then multiply that again by cos^2x/cos^2x

to get 4cos^2x/1+sin^2x+cos^2x

in your example where did the +1 go

Where did the 1+ go on the bottom?

?

yeah agreed so for my information in the denominator

the sin^2x/cos^2x was multiply by cos^2x/cos^2x to leave sin^2x which makes it

4/1+sin^2x

No - what I did was this (leaving the -3 term off for simplicity):


\frac 4 {1 + \frac {\sin^2x}{\cos^2x}} = \frac 4 {1 + \frac {\sin^2x}{\cos^2x}} \times \frac {\cos^2x}{\cos^2x} = \frac {4 \cos^2x}{(1+ \frac {\sin^2x}{\cos^2x}) \cos^2x} = \frac {4 \cos^2x}{\cos^2x + \frac {\sin^2x}{\cancel{\cos^2x}} \cancel{ \cos^2x}} = \frac {4 \cos^2x}{\cos^2x + \sin^2x} = 4 \cos^2x


I hope this is clear.